Chapter 14, Problem 14.107SP

(a)

Interpretation Introduction

Interpretation:

The rate constant of the given reaction has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

$k=A{e}^{-E_a/RT}$

• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of 8.314 J/K.mol
• T represents the absolute temperature
• A represents the frequency factor or collision frequency
• e represents the base of natural logarithm
•  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
• The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Answer to Problem 14.107SP

The rate constant of the given reaction is $k=0.0350{\min }^{-1}$

Explanation of Solution

We can easily determine the rate constant of a reaction if half-life value is given by using half-life formula.

$k=\frac{0.693}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$k=\frac{0.693}{19.8\min }$

$k=0.0350{\min }^{-1}$

(b)

Interpretation Introduction

Interpretation:

The activation energy for the decomposition of benzoyl peroxide has to be calculated.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

$k=A{e}^{-E_a/RT}$

• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of 8.314 J/K.mol
• T represents the absolute temperature
• A represents the frequency factor or collision frequency
• e represents the base of natural logarithm
•  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
• The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Answer to Problem 14.107SP

The activation energy for the decomposition of benzoyl peroxide is ${\text{E}}_{\text{a}}\text{=1}{\text{.1×10}}^{\text{5}}\text{J/mol = 110}\text{kJ/mol}$

Explanation of Solution

The activation energy can be calculated by using modified Arrhenius equation and it can represented as follows

$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Where, ${\text{k}}_{\text{1}}$ value is determined in step (a) ${\text{k}}_{\text{1}}=0.0350{\min }^{-1}$

Now, the given rate constant at $70°C$ is ${\text{k}}_2=1.58\times {10}^{-3}{\min }^{-1}$

Substitute the given values in the above said equation to get activation energy for the

$\text{ln}\frac{\text{0}\text{.0350}{\text{min}}^{\text{-1}}}{\text{1}{\text{.58×10}}^{\text{-3}}{\text{min}}^{\text{-1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\left(\text{8}\text{.314J/K}\text{.mol}\right)}\left[\frac{\text{1}}{\text{343K}}\text{-}\frac{\text{1}}{\text{373K}}\right]$

${\text{E}}_{\text{a}}\text{=1}{\text{.1×10}}^{\text{5}}\text{J/mol = 110}\text{kJ/mol}$

(c)

Interpretation Introduction

Interpretation:

The reactant, product, and intermediate form the given elementary steps have to be written.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

$k=A{e}^{-E_a/RT}$

• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of 8.314 J/K.mol
• T represents the absolute temperature
• A represents the frequency factor or collision frequency
• e represents the base of natural logarithm
•  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
• The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Answer to Problem 14.107SP

The reactant, product, and intermediate form the given elementary steps are written.

Explanation of Solution

The given steps are all elementary steps, the rate law for the given steps can be deduced simply as follows.

$\begin{array}{l}{\text{Initiation: rate = k}}_{\text{i}}{\text{[R}}_{\text{2}}\text{]}\\ {\text{Propagation: rate = k}}_{\text{p}}{\text{[M][M}}_{\text{1}}\text{]}\\ {\text{Termination: rate = k}}_{\text{t}}\text{[M'][M'']}\end{array}$

• Reactants are: ethylene monomers
• Product is: polyethylene
• Intermediate are: M’, M”, and so on

 (R-species also qualifies as an intermediate)

(d)

Interpretation Introduction

Interpretation:

What condition would favour the growth of long, high-molar-mass polyethylenes has to be explained.

Concept introduction:

Half-life: The time required for half of a reactant to be consumed in a reaction is said to be half-life.

• Half-life of a reaction is represented by the symbol as $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• Half-life is discovered by Ernest Rutherford's in 1907 from the original term half-life period.
• The half-life period is then shortened as half-life in early 1950s.

Arrhenius equation:

• Arrhenius equation is a formula that represents the temperature dependence of reaction rates
• The Arrhenius equation has to be represented as follows

$k=A{e}^{-E_a/RT}$

• $E_a$ represents the activation energy and it’s unit is kJ/mol
• R represents the universal gas constant and it has the value of 8.314 J/K.mol
• T represents the absolute temperature
• A represents the frequency factor or collision frequency
• e represents the base of natural logarithm
•  Arrhenius equation equation was proposed by Svante Arrhenius in 1889.
• The activation energy for the given decomposition can be calculated from the modified Arrhenius equation.

$\ln \frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Reaction: Substances which are mutually involved each other in a chemical process and changed into different substances.

Answer to Problem 14.107SP

What condition would favour the growth of long, high-molar-mass polyethylenes is explained.

Explanation of Solution

• A high rate of propagations and a low rate of termination will favour the growth of long polymers.
• The rate law of propagation depends on the concentration of ethylene monomer, when we increase the concentration of ethylene the rate of the propagation also increases.
• The rate law of termination shows that the low concentration of the radical fragment M’ or M” and is lead to slower the rate of termination.  Which is accomplished by taking a low concentration of the initiator, ${\text{R}}_{\text{2}}$